Recipe

Classify a candidate H against G

Use when: you're given a graph G and a candidate H, or just a vertex subset, and asked whether H is a subgraph, an induced subgraph, a spanning subgraph, or none of these

Goal: a definite classification of H against G, backed by the exact vertex and edge check that produced it

Taught in Subgraphs, induced subgraphs, and spanning subgraphs

Quick recipe

Confirm H is a legitimate subgraph at all
Check whether H is spanning
Check whether H is induced
Combine the two checks into a final label

1
Check that $V(H) \subseteq V(G)$ , that $E(H) \subseteq E(G)$ , and that every edge in $E(H)$ has both endpoints in $V(H)$ .

Example

Using $G$ with $V(G)=\{A,B,C,D,E\}$ and $E(G) = \{AB,AC,AD,CD,BE,DE\}$ : any $H$ built only from these vertices and only from these six edges passes Step 1 automatically.
- all three conditions hold — continue to Step 2
- any condition fails, such as an edge of $H$ not in $G$ , or an edge whose endpoint is missing from $V(H)$ — $H$ is not a subgraph of $G$ at all, stop here
Why: a subgraph is only ever allowed to drop vertices and edges from $G$ , never to add a vertex, an edge, or an edge with a missing endpoint.
2
Compare $V(H)$ to $V(G)$ directly. Spanning requires every vertex of $G$ to survive in $H$ , with the edge set free to be any subset.

Example

For $H$ with $V(H) = \{A,B,C,D,E\}$ and $E(H) = \{AB, CD\}$ : $V(H) = V(G)$ exactly, so $H$ is spanning.
- $V(H) = V(G)$ exactly — $H$ is at least a candidate spanning subgraph, continue to Step 3
- $V(H)$ is a proper subset of $V(G)$ — $H$ cannot be spanning, but still check Step 3 on this smaller vertex set
Why: spanning is entirely a vertex-side condition, unrelated to how many of $G$ 's eligible edges $H$ actually kept.
3
For the vertex set $V(H)$ , list every $G$ -edge with both endpoints in $V(H)$ ; call this the eligible edge set. Compare it to $E(H)$ exactly.

Example

For $V(H) = \{B,D,E\}$ , the eligible edges among $G$ 's six are $BE$ and $DE$ ( $BD$ is not a $G$ -edge). If $E(H) = \{BE, DE\}$ , $H$ is induced on $\{B,D,E\}$ . Back on $V(H) = V(G)$ , the eligible set is all six $G$ -edges; the earlier $H$ with $E(H) = \{AB,CD\}$ is missing four of them ( $AC$ , $AD$ , $BE$ , $DE$ ), so that $H$ is not induced.
- $E(H)$ equals the full eligible set — $H$ is induced on $V(H)$
- $E(H)$ is missing at least one eligible edge — $H$ is not induced
Why: an induced subgraph is never allowed to selectively drop an edge that both its endpoints qualify for, unlike an ordinary subgraph.
4
Passing both Step 2 and Step 3 only happens when $V(H) = V(G)$ and every eligible edge survived, which forces $H = G$ itself. Spanning-but-not-induced keeps every vertex while dropping some eligible edge. Induced-but-not-spanning restricts to fewer vertices while keeping all of them eligible among those. Failing both restricts the vertices and still drops an eligible edge among the ones that remain.

Why: the two checks are independent, so all four combinations are genuinely possible outcomes, not just two.

Common mistakes

Assuming "every vertex kept" by itself makes a subgraph induced -- spanning and induced are independent checks.
Building the eligible edge set from the wrong vertex set, using $V(G)$ instead of the smaller $V(H)$ actually being tested.
Concluding "not a subgraph" just because $H$ is missing some of $G$ 's edges -- an ordinary subgraph is allowed to drop edges freely.
Forgetting that dropping even one eligible edge disqualifies $H$ as induced, no matter how many other eligible edges it kept.